Asked by jj reddick
the following data were collected when titrating 25.00 mL samples of HCL with 0.511 M NaOH:
initial burette reading (mL): trial 1=1.06 trial2=0.04 trial3=1.03
final burette reading (mL): trial1= 27.60 trial2=26.21 trial3=27.22
initial burette reading (mL): trial 1=1.06 trial2=0.04 trial3=1.03
final burette reading (mL): trial1= 27.60 trial2=26.21 trial3=27.22
Answers
Answered by
bobpursley
MolarityAcid*.025=.511M*volumebase
averaging the volume of base:
27.60-1.06=26.54
26.21-.04=26.17
27.22-1.03=26.18
Now I am wondering what your question is.
averaging the volume of base:
27.60-1.06=26.54
26.21-.04=26.17
27.22-1.03=26.18
Now I am wondering what your question is.
Answered by
jj reddick
using this data what is the [HCL]?
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